Escreva em somatório as somas indicadas. C sum _(i=1)^3(2i+1) d) sum _(k=0)^4(k^2-1) e) sum _(n=1)^5((2n)/(n+1)) f) sum _(n=1)^4((n!)/(n+2)) g sum _(k=1)^3(k^2ln(k))

C) $\sum _{i=1}^{3}(2i+1) = (2 \cdot 1 + 1) + (2 \cdot 2 + 1) + (2 \cdot 3 + 1) = 2 + 5 + 7 = 14$<br /><br />D) $\sum _{k=0}^{4}(k^{2}-1) = (0^2 - 1) + (1^2 - 1) + (2^2 - 1) + (3^2 - 1) + (4^2 - 1) = -1 + 0 + 3 + 8 + 15 = 25$<br /><br />E) $\sum _{n=1}^{5}(\frac {2n}{n+1}) = \frac{2 \cdot 1}{1 + 1} + \frac{2 \cdot 2}{2 + 1} + \frac{2 \cdot 3}{3 + 1} + \frac{2 \cdot 4}{4 + 1} + \frac{2 \cdot 5}{5 + 1} = \frac{2}{2} + \frac{4}{3} + \frac{6}{4} + \frac{8}{5} + \frac{10}{6} = 1 + \frac{4}{3} + \frac{3}{2} + \frac{8}{5} + \frac{5}{3} = \frac{15}{15} + \frac{20}{15} + \frac{22.5}{15} + \frac{24}{15} + \frac{25}{15} = \frac{106.5}{15} = 7.1$<br /><br />F) $\sum _{n=1}^{4}(\frac {n!}{n+2}) = \frac{1!}{1+2} + \frac{2!}{2+2} + \frac{3!}{3+2} + \frac{4!}{4+2} = \frac{1}{3} + \frac{2}{4} + \frac{6}{5} + \frac{24}{6} = \frac{1}{3} + \frac{1}{2} + \frac{6}{5} + 4 = \frac{10}{30} + \frac{15}{30} + \frac{36}{30} + \frac{120}{30} = \frac{181}{30} = 6.03$<br /><br />G) $\sum _{k=1}^{3}(k^{2}ln(k)) = 1^2 \cdot ln(1) + 2^2 \cdot ln(2) + 3^2 \cdot ln(3) = 0 + 4 \cdot ln(2) + 9 \cdot ln(3) = 4 \cdot ln(2) + 9 \cdot ln(3) = 4 \cdot 0.6931 + 9 \cdot 1.0986 = 2.7724 + 9.8829 = 12.6553$